Lable point B at (2,-3/2)

New shoes are on SALE. You find a pair you like for $85 dollars. But you only have $45 with you. So, you pay $40 and

andrew11 [14]

Answer:

C) $46.80

Explanation:

The price of new shoes = $85

The amount paid for the shoes = $ 40

The balance amount for the shoes = $85 - $40 = $ 45

The rate of interest = 8%

The time period = 6 months = 0.5 years

Simple interest:

Simple interest = = $1.8

So, Amount = Principal + Interest

or, Amount = $45 + $1.8 = $ 46.8

The balance amount owned in six months is $46.8

3 0

2 years ago

Solve equation by using the quadratic formula. 2x2 + 13x = 0

Nadusha1986 [10]

THE answer is 17 because 2X2= 4+13= 17
Your Welcome :D

5 0

3 years ago

Read 2 more answers

The side of cube is 3m 30cm long, find out its volume

Anettt [7]

90 volume because 3 multiples by 30 = 90

7 0

2 years ago

Maria and Nate graphed lines on a coordinate plane. Maria's line is represented by the equation y = -5/6x +8. Nate's line is per

eduard

Answer:

c) 6x - 5y = 15

Step-by-step explanation:

Slope-intercept form of a linear equation: $y=mx+b$

(where m is the slope and b is the y-intercept)

Maria's line: $y=-\dfrac{5}{6}x+8$

Therefore, the slope of Maria's line is $-\frac{5}{6}$

If two lines are perpendicular to each other, the product of their slopes will be -1.

Therefore, the slope of Nate's line (m) is:

$\begin{aligned}\implies m \times -\dfrac{5}{6} &=-1\\m & =\dfrac{6}{5}\end{aligned}$

Therefore, the linear equation of Nate's line is:

$y=\dfrac{6}{5}x+b\quad\textsf{(where b is some constant)}$

Rearranging this to standard form:

$\implies y=\dfrac{6}{5}x+b$

$\implies 5y=6x+5b$

$\implies 6x-5y=-5b$

Therefore, option c could be an equation for Nate's line.

3 0

2 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago